Use Newton's method to find the first 5 approximations of the solution to the equation $LaTeX: \displaystyle \cos{\left(x \right)}= \frac{331 x^{3}}{1000} - 6$ using $LaTeX: \displaystyle x_0=3$ .

Using the formula for Newton's method gives $LaTeX: x_{n+1} = x_{n} - \frac{- \frac{331 x_{n}^{3}}{1000} + \cos{\left(x_{n} \right)} + 6}{- \frac{993 x_{n}^{2}}{1000} - \sin{\left(x_{n} \right)}}$ Using $LaTeX: \displaystyle x_0 = 3$ and $LaTeX: \displaystyle n = 0,1,2,3,$ and $LaTeX: \displaystyle 4$ gives: $LaTeX: x_{1} = (3.0000000000) - \frac{- \frac{331 (3.0000000000)^{3}}{1000} + \cos{\left((3.0000000000) \right)} + 6}{- \frac{993 (3.0000000000)^{2}}{1000} - \sin{\left((3.0000000000) \right)}} = 2.5674222754$ $LaTeX: x_{2} = (2.5674222754) - \frac{- \frac{331 (2.5674222754)^{3}}{1000} + \cos{\left((2.5674222754) \right)} + 6}{- \frac{993 (2.5674222754)^{2}}{1000} - \sin{\left((2.5674222754) \right)}} = 2.5051616550$ $LaTeX: x_{3} = (2.5051616550) - \frac{- \frac{331 (2.5051616550)^{3}}{1000} + \cos{\left((2.5051616550) \right)} + 6}{- \frac{993 (2.5051616550)^{2}}{1000} - \sin{\left((2.5051616550) \right)}} = 2.5039607374$ $LaTeX: x_{4} = (2.5039607374) - \frac{- \frac{331 (2.5039607374)^{3}}{1000} + \cos{\left((2.5039607374) \right)} + 6}{- \frac{993 (2.5039607374)^{2}}{1000} - \sin{\left((2.5039607374) \right)}} = 2.5039602965$ $LaTeX: x_{5} = (2.5039602965) - \frac{- \frac{331 (2.5039602965)^{3}}{1000} + \cos{\left((2.5039602965) \right)} + 6}{- \frac{993 (2.5039602965)^{2}}{1000} - \sin{\left((2.5039602965) \right)}} = 2.5039602965$